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Math Help - Evaluating integrals

  1. #1
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    Evaluating integrals

    I have a final monday and have been working on this integral and cant figure it out.

    evaluate integral 1/(e^(3x)-e^(x))
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  2. #2
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    Set u=e^x. Then you get a rational function \frac{1}{u(u-1)(u+1)}. Rewrite this rational function as a sum of rational functions \frac{A}{u}+\frac{B}{u+1}+\frac{C}{u-1}, for some constants A,B,C. Then integrate.
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  3. #3
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    All you need to do is to see that  e^{3x}  = (e^{x})^{3}

    Sub in  u = e^{x}  , du = e^{x}

    Edit

    vemrygh beat me to it.
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  4. #4
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    \frac{1}{u(u-1)(u+1)}=\frac{1}{u\left( u^{2}-1 \right)}=\frac{u^{2}-\left( u^{2}-1 \right)}{u\left( u^{2}-1 \right)}=\frac{u}{u^{2}-1}-\frac{1}{u}.
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  5. #5
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     \frac{1}{e^{3x}-e^{x}} =
    \frac{-2e^{2x} + 2 + e^{3x} + e^{2x} - e^{3x} + e^{2x}}{2(e^{3x}-e^{x})}=
    \frac{2(e^x-1)(e^x+1)+(e^x)(e^x+1))(e^x)-(e^x)(e^x-1))(e^x)}{2*(e^x)(e^x-1)(e^x+1)}=<br />
    -\frac{2(e^x-1)(e^x +1)}{2(e^x)(e^x-1)(e^x +1)} +\frac{(e^x)(e^x +1)(e^x)}{2(e^x)(e^x-1)(e^x +1)} -\frac{(e^x)(e^x-1)(e^x)}{2(e^x)(e^x-1)(e^x +1)}=
    -\frac{1}{e^x} + \frac{e^x(e^x+1)}{2(e^x-1)(e^x+1)} - \frac{e^x(e^x-1)}{2(e^x+1)(e^x-1)}=
    -\frac{1}{e^x} + \frac{e^x}{2(e^x-1)} - \frac{e^x}{2(e^x+1)}

    which can easily be integrated to:
    \frac{1}{e^x} + \frac{1}{2}ln(e^x-1)- \frac{1}{2}ln(e^x+1)
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